Let $L$ be some language.
The closure of a language $L$ is denoted $L^*$ and represents the set of those strings that can be formed by taking any number of strings from $L$, possibly with repetitions (i.e., the same string may be selected more than once) and concatenating all of them.
So, a string $w$ belongs to $L^*$ iff $w$ can be broken into a sequence of substrings such that each substring belongs to $L.$
In more Formal Way, a string $w$ belongs to $L^*$ iff there is some integer $k \geq 0, $ such that $w = w_1w_2w_3 \dots w_k$ and each $w_i \in L.$
Moreover, a string $w$ belongs to $L^n$ iff $w$ can be broken into a sequence of $n$ substrings such that each substring belongs to $L.$
For example, a string $w$ belongs to $L^4$ iff $w$ can be broken into a sequence of $4$ substrings such that each substring belongs to $L.$
Now, we just need to check each string given in the question & see if we can break them into a sequence of substrings such that each substring belongs to $L.$
- $w_1 = \text{abaabaaabaa =}$ $\text{ab aa baa ab aa }$ (This is the Unique breakup of $w_1$. So, $w_1 \in L^5$, Hence, $w_1 \in L^*$. BUT $w_1 \notin L^4$.)
- $w_2 = \text{aaaabaaaa =}$ $\text{aa aa baa aa }$ (This is the Unique breakup of $w_2$. So, $w_2 \in L^4$, Hence, $w_2 \in L^*$. BUT $w_2 \notin L^5$.)
- $w_3 = \text{baaaaabaaaab =}$ $\text{baa aa ab aa aa } \color{red}{b}$ (Since we can’t break $w_3$ into a desired sequence of substrings, hence, $w_3 \notin L^*.$)
- $w_4 = \text{baaaaabaa =}$ $\text{baa aa ab aa }$ (This is the Unique breakup of $w_4$. So, $w_4 \in L^4$, Hence, $w_4 \in L^*$. BUT $w_4 \notin L^5$.)
NOTE: Since language $L$ is prefix-free (i.e. no string of $L$ is prefix of any other string of $L$), we will ALWAYS get an Unique breakup of every string $w \in L^*.$
$\color{red}{\text{Detailed Video Solution, with Complete Analysis:}}$ https://www.youtube.com/watch?v=PjoTWbeUygM&list=PLIPZ2_p3RNHhXeEdbXsi34ePvUjL8I-Q9&index=2
